University Physics V · topic UPV-19.02
Time-translation invariance, Q values, and reaction thresholds
Connect the ideas, make your working visible, and test the result. Use the hints when you need a starting point; open the full solution after your attempt.
Further practice
Work through a fresh case before tackling the extension.
Energy conservation, Q values and thresholds: predict a finite change
Energy is conserved whenever H has no explicit time dependence, because H generates time translation and trivially commutes with itself; the invariant s = (Σp)² then decides whether the conserved energy is enough, since a reaction needs √s ≥ Σm_f c² in every frame. Use the supplied topic model Y_out = dm*(0.938+mb+dm/2)/max(mb,0.01), where Y_out is THRESHOLD T_thr (GeV). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; max(a,b) selects the larger value. mb is Target mass m_b (at rest), measured in GeV. Keep tb (Beam kinetic energy T) = 3 GeV; dm (Mass created ΔM = Σm_f − m_p − m_b) = 1.876 GeV fixed. For this exercise use only 1.15709 GeV ≤ mb ≤ 2.78091 GeV on this continuous branch. A grid of test settings is an input restriction for this exercise, not by itself a claim of physical quantisation. This calculation retains the topic model's stated assumptions; it does not extend them to a different physical setup. Calculate Y_out at mb = 1.51433 GeV and 1.74166 GeV. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δmb across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
THRESHOLD T_thr against mb
Predictions from the supplied Energy conservation, Q values and thresholds model, with all other inputs fixed as stated in the question. 5 calibration points support a trend comparison; they are not measured data or a straight-line law.
Vertical axis (y): THRESHOLD T_thr / GeV
Horizontal axis (x): mb: Target mass m_b (at rest) / GeV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| mb: Target mass m_b (at rest) / GeV | THRESHOLD T_thr / GeV |
|---|---|
| 1.1570875 | 4.917581557142393 |
| 1.56304375 | 4.127617077257114 |
| 1.9690000000000003 | 3.6633925850685616 |
| 2.3749562500000003 | 3.357869823917808 |
| 2.7809125000000003 | 3.1415471899960887 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of mb separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At mb = 1.51433 GeV, substitute into Y_out = dm*(0.938+mb+dm/2)/max(mb,0.01): Y₀ = 4.20005 GeV.
- At mb = 1.74166 GeV, the same fixed-parameter model gives Y₁ = 3.8967 GeV.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.303353 GeV; the input change is Δmb = 0.227336 GeV.
- Average sensitivity is ΔY_out/Δmb = -1.33438 GeV per GeV. Its negative sign means that this response decreases over the specified interval.
- This is a secant slope between two settings. It equals a point derivative only for a locally linear relation or in a justified small-interval limit.
Answer Y₀ = 4.20005 GeV, Y₁ = 3.8967 GeV; ΔY_out = -0.303353 GeV; average sensitivity = -1.33438 GeV per GeV.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Extra challenging questions
Work through the linked stages and explain what each result means.
Energy conservation, Q values and thresholds: infer an input with uncertainty
Energy is conserved whenever H has no explicit time dependence, because H generates time translation and trivially commutes with itself; the invariant s = (Σp)² then decides whether the conserved energy is enough, since a reaction needs √s ≥ Σm_f c² in every frame. Use the supplied topic model Y_out = dm*(0.938+mb+dm/2)/max(mb,0.01), where Y_out is THRESHOLD T_thr (GeV). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; max(a,b) selects the larger value. mb is Target mass m_b (at rest), measured in GeV. Keep tb (Beam kinetic energy T) = 3 GeV; dm (Mass created ΔM = Σm_f − m_p − m_b) = 1.876 GeV fixed. For this exercise use only 1.15709 GeV ≤ mb ≤ 2.78091 GeV on this continuous branch. A grid of test settings is an input restriction for this exercise, not by itself a claim of physical quantisation. This calculation retains the topic model's stated assumptions; it does not extend them to a different physical setup. A reading gives Y_out = 3.5147325 GeV, with a bounded reading uncertainty ±0.055978 GeV. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown mb by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000162383 GeV, using a numerical bracket if required. Find the full input interval consistent with the bounded reading uncertainty and compare it with a symmetric input-error assumption. Do not assume that the supplied calibration points prove monotonicity outside this branch.
THRESHOLD T_thr against mb
Predictions from the supplied Energy conservation, Q values and thresholds model, with all other inputs fixed as stated in the question. 5 calibration points support a trend comparison; they are not measured data or a straight-line law.
Vertical axis (y): THRESHOLD T_thr / GeV
Horizontal axis (x): mb: Target mass m_b (at rest) / GeV
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| mb: Target mass m_b (at rest) / GeV | THRESHOLD T_thr / GeV |
|---|---|
| 1.1570875 | 4.917581557142393 |
| 1.56304375 | 4.127617077257114 |
| 1.9690000000000003 | 3.6633925850685616 |
| 2.3749562500000003 | 3.357869823917808 |
| 2.7809125000000003 | 3.1415471899960887 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
The model's endpoint predictions bracket the reading. Solve Y_out(mb) = Y_measured and then solve again at Y_measured − u and Y_measured + u; a nonlinear inverse can give unequal input deviations.
Reveal the worked solution
- Endpoint predictions are Y_out(1.15709) = 4.91758 GeV and Y_out(2.78091) = 3.14155 GeV. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives mb_line = 1.15709 + (3.51473 − 4.91758)(2.78091 − 1.15709)/(3.14155 − 4.91758) = 2.43971 GeV. This estimates the inverse of the endpoint chord, rather than assuming that chord is the physical model.
- Use the actual equation, not the chord, and bracket its root. Evaluating the midpoint and retaining the sign-changing half-interval gives mb = 2.14762 GeV; substituting back gives Y_out = 3.5147325 GeV.
- The interpolation error in the input is |mb_line − mb| = 0.292089 GeV. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 3.4587545 GeV and 3.5707105 GeV. Sort the roots even if the response decreases: 2.07668 GeV ≤ mb ≤ 2.22358 GeV.
- The lower and upper input deviations are 0.0709381 GeV and 0.0759559 GeV. They differ because the inverse response is nonlinear; a symmetric error bar would only be a local approximation. This interval includes reading uncertainty only, not model error or uncertainty in fixed inputs.
Answer mb ≈ 2.14762 GeV; compatible interval [2.07668, 2.22358] GeV. Endpoint interpolation gives 2.43971 GeV.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.